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;
const int N = 121111;
int num[N];
struct SegTree {
  int stl[4 * N], str[4 * N];
  long long stsum[4 * N];
  int stmax[4 * N];
  inline SegTree() {}
  inline void build(int left, int right, int x = 1) {
    stl[x] = left;
    str[x] = right;
    if (left + 1 == right) {
      stmax[x] = left;
      stsum[x] = num[left];
    } else {
      int mid = (left + right) >> 1;
      build(left, mid, 2 * x);
      build(mid, right, 2 * x + 1);
      stsum[x] = stsum[2 * x] + stsum[2 * x + 1];
      stmax[x] = num[stmax[2 * x]] > num[stmax[2 * x + 1]] ? stmax[2 * x]
                                                           : stmax[2 * x + 1];
    }
  }
  inline long long qsum(int left, int right, int x = 1) {
    if (left == stl[x] && right == str[x]) return stsum[x];
    int mid = (stl[x] + str[x]) >> 1;
    if (right <= mid)
      return qsum(left, right, 2 * x);
    else if (left >= mid)
      return qsum(left, right, 2 * x + 1);
    else
      return qsum(left, mid, 2 * x) + qsum(mid, right, 2 * x + 1);
  }
  inline int qmax(int left, int right, int x = 1) {
    if (left == stl[x] && right == str[x]) return stmax[x];
    int mid = (stl[x] + str[x]) >> 1;
    if (right <= mid)
      return qmax(left, right, 2 * x);
    else if (left >= mid)
      return qmax(left, right, 2 * x + 1);
    else {
      int ql = qmax(left, mid, 2 * x);
      int qr = qmax(mid, right, 2 * x + 1);
      if (num[ql] > num[qr])
        return ql;
      else
        return qr;
    }
  }
  inline void update(int pos, int x = 1) {
    if (stl[x] + 1 == str[x])
      stsum[x] = num[pos];
    else {
      int mid = (stl[x] + str[x]) >> 1;
      if (pos < mid)
        update(pos, 2 * x);
      else
        update(pos, 2 * x + 1);
      stsum[x] = stsum[2 * x] + stsum[2 * x + 1];
      stmax[x] = num[stmax[2 * x]] > num[stmax[2 * x + 1]] ? stmax[2 * x]
                                                           : stmax[2 * x + 1];
    }
  }
} A;
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> num[i];
  A.build(1, n + 1);
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      cout << A.qsum(l, r + 1) << '\n';
    } else if (type == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      bool flag = false;
      do {
        flag = false;
        int t = A.qmax(l, r + 1);
        if (num[t] >= x) flag = true;
        num[t] %= x;
        A.update(t);
      } while (flag);
    } else {
      int k, x;
      cin >> k >> x;
      num[k] = x;
      A.update(k);
    }
  }
  return 0;
}