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 double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
inline long long power(long long bs, long long k) {
  long long x = 1LL, y = bs;
  if (k == 0) return 1LL;
  while (k > 0) {
    if (k % 2) x *= y;
    y *= y;
    k /= 2;
  }
  return x;
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
string to_str(T t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
long long a[100100], n, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.unsetf(ios::floatfield);
  cout.precision(10);
  cout.setf(ios::fixed, ios::floatfield);
  ;
  long long i, j, l, r, k;
  cin >> n >> m;
  for (i = 0; i < n; ++i) cin >> a[i];
  vector<vector<long long> > ncr(n + 110, vector<long long>(111, 0));
  vector<vector<long long> > cnt(n + 110, vector<long long>(111, 0));
  for (i = 0; i < n + 110; ++i) {
    ncr[i][0] = 1LL;
    k = min(i, 109LL);
    for (j = 1; j <= k; ++j) {
      ncr[i][j] = ncr[i - 1][j] + ncr[i - 1][j - 1];
      ncr[i][j] %= 1000000007;
    }
  }
  vector<vector<pair<long long, long long> > > sub;
  sub.resize(n + 10);
  for (i = 0; i < m; ++i) {
    cin >> l >> r >> k;
    cnt[l - 1][k]++;
    sub[r].emplace_back(pair<long long, long long>(k, l - 1));
  }
  vector<long long> x(110, 0);
  for (i = 0; i < n; ++i) {
    vector<long long> y = cnt[i];
    long long ad = 0LL;
    for (j = 100; j >= 0; --j) {
      ad = (x[j] + ad) % 1000000007;
      y[j] = (y[j] + ad) % 1000000007;
    }
    x = y;
    long long sz = (long long)sub[i].size();
    for (long long p = 0; p < sz; ++p) {
      for (j = 0; j <= sub[i][p].first; ++j) {
        l = i - sub[i][p].second - 1 + sub[i][p].first - j;
        r = sub[i][p].first - j;
        x[j] -= ncr[l][r];
        x[j] %= 1000000007;
      }
    }
    for (j = 100; j >= 0; --j) a[i] = (a[i] + x[j] + 1000000007) % 1000000007;
    cout << a[i] << '\n';
  }
  return 0;
}