CodeForces

Discription

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you qqueries. In the i-th query you should compute and print the ugliness of a substring(continuous subsequence) of s starting at the index a_i and ending at the index b_i(inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8

Output

4
3
-1

Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15

Output

-1
2
1
-1
-1

Input

4 2
1234
2 4
1 2

Output

-1
-1

Note

In the first sample:

• In the first query, ugliness("20166766") = 4 because all four sixes must be removed.
• In the second query, ugliness("2016676") = 3 because all three sixes must be removed.
• In the third query, ugliness("0166766") =  - 1 because it's impossible to remove some digits to get a nice string.

In the second sample:

• In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice.
• In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170".

     建一个节点和字符集很少的有限状态自动机，然后对于线段树每个区间用一个矩阵记录 从一个状态 到 另一个状态的 最小代价。

    因为这个矩阵是满足结合律的，所以直接做就行了。

#include<bits/stdc++.h>
#define ll long long
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mid (l+r>>1)
using namespace std;
const int maxn=200005;
struct node{
	int a[5][5];
	inline void init(){ memset(a,0x3f,sizeof(a));}
	node operator *(const node &u)const{
		node r; r.init();
		for(int k=0;k<5;k++)
		    for(int i=0;i<5;i++)
		        for(int j=0;j<5;j++) r.a[i][j]=min(r.a[i][j],a[i][k]+u.a[k][j]);
		return r;
	}
}S[maxn*4+5],ANS;
int n,m,le,ri;
char s[maxn];

inline void Set(node &x,int y){
	x.init(); for(int i=0;i<5;i++) x.a[i][i]=0;
	if(y==2) x.a[0][0]=1,x.a[0][1]=0;
	else if(!y) x.a[1][1]=1,x.a[1][2]=0;
	else if(y==1) x.a[2][2]=1,x.a[2][3]=0;
	else if(y==7) x.a[3][3]=1,x.a[3][4]=0;
	else if(y==6) x.a[3][3]=1,x.a[4][4]=1;
}

void build(int o,int l,int r){
	if(l==r){ Set(S[o],s[l]-'0'); return;}
	build(lc,l,mid),build(rc,mid+1,r);
	S[o]=S[lc]*S[rc];
}

void query(int o,int l,int r){
	if(l>=le&&r<=ri){ ANS=ANS*S[o]; return;}
	if(le<=mid) query(lc,l,mid);
	if(ri>mid) query(rc,mid+1,r);
}

inline void solve(){
	build(1,1,n);
	while(m--){
		ANS.init(); for(int i=0;i<5;i++) ANS.a[i][i]=0;
		scanf("%d%d",&le,&ri),query(1,1,n);
		printf("%d
",ANS.a[0][4]<=n?ANS.a[0][4]:-1);
	}
}

int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	solve();
	return 0;
}