CodeForces 408E Curious Array(组合数学+差分)

You've got an array consisting of n integers: a[1], a[2], ..., a[n]. Moreover, there are m queries, each query can be described by three integers l_i, r_i, k_i. Query l_i, r_i, k_i means that we should add  to each element a[j], where l_i ≤ j ≤ r_i.

Record  means the binomial coefficient, or the number of combinations from yelements into groups of x elements.

You need to fulfil consecutively all queries and then print the final array.

Input

The first line contains integers n, m (1 ≤ n, m ≤ 10⁵).

The second line contains n integers a[1], a[2], ..., a[n] (0 ≤ a_i ≤ 10⁹) — the initial array.

Next m lines contain queries in the format l_i, r_i, k_i — to all elements of the segment l_i... r_i add number  (1 ≤ l_i ≤ r_i ≤ n; 0 ≤ k ≤ 100).

Output

Print n integers: the i-th number is the value of element a[i] after all the queries. As the values can be rather large, print them modulo 1000000007 (10⁹ + 7).

Examples

Input

5 1
0 0 0 0 0
1 5 0

Output

1 1 1 1 1

Input

10 2
1 2 3 4 5 0 0 0 0 0
1 6 1
6 10 2

Output

2 4 6 8 10 7 3 6 10 15

题意：给出n个数，m次操作
每次操作给出li，ri，ki
将li-ri的范围里的第j个数加上 

题解：考虑到上面的ki是在每个操作里是不会变的
考虑在杨辉三角中找规律

k=i是k=i-1的前缀和
然后可以考虑差分
在高维进行好差分后一维一维的降下来
注意差分的时候要每一层都差分

代码如下:

#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mod 1000000007
using namespace std;

long long fac[200010],inv[200010],n,m,a[110][100010],sum[110][100010];

long long kasumi(long long a,long long b)
{
    long long ans=1ll;
    while(b)
    {
        if(b&1)
        {
            ans=ans*a%mod;
        }
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}

long long c(long long a,long long b)
{
    return fac[a]*inv[b]%mod*inv[a-b]%mod;
}

int main()
{
    fac[0]=1,inv[0]=1;
    for(int i=1; i<=200000; i++)
    {
        fac[i]=fac[i-1]*i*1ll%mod;
    }
    inv[200000]=kasumi(fac[200000],mod-2);
    for(int i=199999; i>=1; i--)
    {
        inv[i]=inv[i+1]*(i+1)*1ll%mod;
    }
    scanf("%lld%lld",&n,&m);
    for(int i=0; i<n; i++)
    {
        scanf("%lld",&a[0][i]);
    }
    long long l,r,k;
    while(m--)
    {
        scanf("%lld%lld%lld",&l,&r,&k);
        l--,k++;
        a[k][l]++;
        for(int i=1; i<=k; i++)
        {
            a[i][r]-=c(k-i+r-l-1,k-i);
            a[i][r]=(a[i][r]+mod)%mod;
        }
    }
    for(int k=108; k>=0; k--)
    {
        for(int i=0;i<n;i++)
        {
            a[k][i]+=sum[k+1][i+1];
            a[k][i]%=mod;
        }
        for(int i=0;i<n;i++)
        {
            sum[k][i+1]+=sum[k][i]+a[k][i];
            sum[k][i+1]%=mod;
        }
    }
    for(int i=0;i<n;i++)
    {
        printf("%lld ",a[0][i]%mod);
    }
}