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 maxn = 100 * 100 * 10 + 10;
pair<long long, long long> maxx[4 * maxn];
long long sum[4 * maxn], a[maxn];
pair<long long, long long> getmax(int l, int r, int ind, int b, int e) {
  pair<long long, long long> ret1, ret2, ret3;
  ret1 = ret2 = ret3 = make_pair(0, 0);
  if (l <= b && e <= r) {
    return maxx[ind];
  }
  int mid = (b + e) / 2;
  if (l < mid) {
    ret1 = getmax(l, r, ind * 2, b, mid);
  }
  if (r > mid) {
    ret2 = getmax(l, r, ind * 2 + 1, mid, e);
  }
  ret3 = max(ret1, ret2);
  return ret3;
}
void make(int ind, int b, int e) {
  if (e - b == 1) {
    sum[ind] = a[b];
    maxx[ind].first = a[b];
    maxx[ind].second = b;
    return;
  }
  int mid = (b + e) / 2;
  make(ind * 2, b, mid);
  make(ind * 2 + 1, mid, e);
  maxx[ind] = max(maxx[ind * 2], maxx[ind * 2 + 1]);
  sum[ind] = sum[ind * 2] + sum[ind * 2 + 1];
  return;
}
void change(int p, long long val, int ind, int b, int e) {
  if (e - b == 1) {
    sum[ind] = val;
    maxx[ind].first = val;
    return;
  }
  int mid = (b + e) / 2;
  if (p < mid)
    change(p, val, ind * 2, b, mid);
  else
    change(p, val, ind * 2 + 1, mid, e);
  maxx[ind] = max(maxx[ind * 2], maxx[ind * 2 + 1]);
  sum[ind] = sum[ind * 2] + sum[ind * 2 + 1];
  return;
}
long long get(int l, int r, int ind, int b, int e) {
  long long ret = 0;
  if (l <= b && e <= r) {
    return sum[ind];
  }
  int mid = (b + e) / 2;
  if (l < mid) ret += get(l, r, ind * 2, b, mid);
  if (r > mid) ret += get(l, r, ind * 2 + 1, mid, e);
  sum[ind] = sum[ind * 2] + sum[ind * 2 + 1];
  return ret;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, m, type, u, k, x, t, l, r;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  make(1, 0, n);
  for (int i = 0; i < m; i++) {
    cin >> type;
    if (type == 1) {
      cin >> l >> r;
      cout << get(l - 1, r, 1, 0, n) << endl;
    } else if (type == 2) {
      cin >> l >> r >> x;
      l--;
      while (getmax(l, r, 1, 0, n).first >= x) {
        u = getmax(l, r, 1, 0, n).second;
        change(u, getmax(l, r, 1, 0, n).first % x, 1, 0, n);
      }
    } else {
      cin >> k >> x;
      k--;
      change(k, x, 1, 0, n);
    }
  }
  return 0;
}