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;
int n, m, x, y, z, top;
long long c[101000][200], b[101000], P = 1e9 + 7, a[101000];
vector<pair<int, pair<int, int> > > q;
int main() {
  scanf("%d%d", &n, &m);
  c[0][0] = 1;
  for (int i = 1; i <= n + 200; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= 200; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % P;
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    q.push_back(make_pair(z, make_pair(y - x, y)));
  }
  sort(q.begin(), q.end());
  top = m - 1;
  for (int i = 100; i >= 0; i--) {
    while (top >= 0 && q[top].first >= i)
      b[q[top].second.second - q[top].second.first]++, top--;
    for (int j = top + 1; j < m; j++)
      b[q[j].second.second + 1] -=
          c[q[j].second.first + q[j].first - i][q[j].first - i],
          b[q[j].second.second + 1] = (b[q[j].second.second + 1] + P) % P;
    for (int i = 2; i <= n; i++) b[i] += b[i - 1], b[i] %= P;
  }
  for (int i = 1; i <= n; i++) printf("%lld ", (a[i] + b[i]) % P);
}