Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class SegmentTree {
 public:
  SegmentTree(const vector<long long int>& value) : value_(value) {
    int size = value.size() * 4;
    sum_.resize(size, 0);
    max_.resize(size, 0);
    index_.resize(size, 0);
    build(1, 0, value.size() - 1);
  }
  long long int get(int i) const { return value_[i]; }
  void assign(int i, long long int x) {
    long long int delta = x - value_[i];
    update(i, delta, 1, 0, value_.size() - 1);
  }
  long long int sum(int L, int R) const {
    return sum(1, 0, value_.size() - 1, L, R);
  }
  int maxIndex(int L, int R) const {
    int index;
    long long int value;
    maxIndex(1, 0, value_.size() - 1, L, R, index, value);
    return index;
  }

 private:
  void build(int id, int l, int r) {
    if (l == r) {
      sum_[id] = value_[l];
      max_[id] = value_[l];
      index_[id] = l;
      return;
    }
    int m = (l + r) / 2;
    build(id * 2, l, m);
    build(id * 2 + 1, m + 1, r);
    pushUp(id);
  }
  void pushUp(int id) {
    int l = id * 2;
    int r = id * 2 + 1;
    sum_[id] = sum_[l] + sum_[r];
    if (max_[l] > max_[r]) {
      max_[id] = max_[l];
      index_[id] = index_[l];
    } else {
      max_[id] = max_[r];
      index_[id] = index_[r];
    }
  }
  void update(int i, long long int delta, int id, int l, int r) {
    if (l == r) {
      value_[i] += delta;
      sum_[id] += delta;
      max_[id] += delta;
      return;
    }
    int m = (l + r) / 2;
    if (i <= m) {
      update(i, delta, id * 2, l, m);
    } else {
      update(i, delta, id * 2 + 1, m + 1, r);
    }
    pushUp(id);
  }
  long long int sum(int id, int l, int r, int L, int R) const {
    if (L <= l && R >= r) {
      return sum_[id];
    }
    int m = (l + r) / 2;
    long long int s = 0;
    if (L <= m) {
      s += sum(id * 2, l, m, L, R);
    }
    if (R > m) {
      s += sum(id * 2 + 1, m + 1, r, L, R);
    }
    return s;
  }
  void maxIndex(int id, int l, int r, int L, int R, int& index,
                long long int& value) const {
    if (L <= l && R >= r) {
      index = index_[id];
      value = max_[id];
      return;
    }
    int m = (l + r) / 2;
    index = L;
    value = value_[L];
    if (L <= m) {
      int li;
      long long int lv;
      maxIndex(id * 2, l, m, L, R, li, lv);
      if (lv > value) {
        index = li;
        value = lv;
      }
    }
    if (R > m) {
      int ri;
      long long int rv;
      maxIndex(id * 2 + 1, m + 1, r, L, R, ri, rv);
      if (rv > value) {
        index = ri;
        value = rv;
      }
    }
  }

 private:
  vector<long long int> value_;
  vector<long long int> sum_;
  vector<long long int> max_;
  vector<int> index_;
};
int main() {
  int n, m;
  cin >> n >> m;
  vector<long long int> value(n);
  for (int i = 0; i < n; ++i) {
    cin >> value[i];
  }
  SegmentTree st(value);
  while (m--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r;
      cin >> l >> r;
      cout << st.sum(l - 1, r - 1) << endl;
    } else if (op == 2) {
      int l, r;
      long long int x;
      cin >> l >> r >> x;
      while (true) {
        int maxIndex = st.maxIndex(l - 1, r - 1);
        if (st.get(maxIndex) < x) {
          break;
        }
        st.assign(maxIndex, st.get(maxIndex) % x);
      }
    } else if (op == 3) {
      int i;
      long long int x;
      cin >> i >> x;
      st.assign(i - 1, x);
    }
  }
  return 0;
}