Problem: 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 li, ri, ki. Query li, ri, ki means that we should add <image> to each element a[j], where li ≤ j ≤ ri.

Record <image> means the binomial coefficient, or the number of combinations from y elements 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 ≤ ai ≤ 109) — the initial array.

Next m lines contain queries in the format li, ri, ki — to all elements of the segment li... ri add number <image> (1 ≤ li ≤ ri ≤ 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
const int maxk = 107;
const long long md = 1e9 + 7;
long long p[maxn][maxk + 1];
long long C[maxn][maxk + 1];
long long A[maxn];
long long B[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int r = (0); r < (maxn); r++) {
    for (int k = (0); k < (maxk); k++) {
      if (k == 0) {
        C[r][k] = 1;
        continue;
      } else if (r == 0) {
        C[r][k] = 0;
        continue;
      } else {
        C[r][k] = (C[r - 1][k] + C[r - 1][k - 1]) % md;
      }
    }
  }
  int n;
  int m;
  cin >> n >> m;
  for (int i = (0); i < (maxn); i++) {
    for (int j = (0); j < (maxk); j++) {
      p[i][j] = 0;
    }
    B[i] = 0;
  }
  for (int i = (1); i <= (n); i++) {
    cin >> A[i];
  }
  while (m--) {
    int l, r, k;
    cin >> l >> r >> k;
    int m = r - l + 1;
    int v = maxk - 1 - k - 1;
    p[l][v] = (p[l][v] + 1) % md;
    p[r + 1][v] = (p[r + 1][v] - 1 + md) % md;
    for (int i = (v + 1); i <= (v + k); i++) {
      p[r + 1][i] = (p[r + 1][i] - C[m + i - v - 1][i - v] + md) % md;
    }
  }
  for (int k = (1); k <= (maxk - 1); k++) {
    p[0][k] = 0;
    for (int i = (0); i < (n + 5); i++) B[i] = 0;
    for (int i = (1); i <= (n); i++) {
      B[i] = (B[i - 1] + p[i][k - 1]) % md;
      p[i][k] = (B[i] + p[i][k]) % md;
    }
  }
  for (int i = (1); i <= (n); i++) {
    A[i] = (p[i][maxk - 1] + A[i]) % md;
    cout << A[i];
    if (i < n) cout << " ";
  }
  cout << endl;
}