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;
long long t = 1;
const long long N = 2e5 + 44;
long long segTree[4 * N], segTree2[4 * N], A[N];
inline long long merge(long long x, long long y) { return x + y; }
inline long long merge2(long long x, long long y) {
  return A[x] > A[y] ? x : y;
}
void update(long long node, long long l, long long r, long long idx,
            long long val) {
  if (l == r) {
    segTree[node] = val;
    segTree2[node] = l;
    A[idx] = val;
    return;
  }
  long long mid = (l + r) / 2;
  if (l <= idx and idx <= mid)
    update(node << 1, l, mid, idx, val);
  else
    update(node << 1 | 1, mid + 1, r, idx, val);
  segTree[node] = merge(segTree[node << 1], segTree[node << 1 | 1]);
  segTree2[node] = merge2(segTree2[node << 1], segTree2[node << 1 | 1]);
}
long long querySum(long long node, long long l, long long r, long long a,
                   long long b) {
  if (l > b || r < a || l > r) return 0;
  if (a <= l and r <= b) return segTree[node];
  long long mid = (l + r) / 2;
  return merge(querySum(node << 1, l, mid, a, b),
               querySum(node << 1 | 1, mid + 1, r, a, b));
}
long long queryMax(long long node, long long l, long long r, long long a,
                   long long b) {
  if (l > b || r < a || l > r) return 0;
  if (a <= l and r <= b) return segTree2[node];
  long long mid = (l + r) / 2;
  return merge2(queryMax(node << 1, l, mid, a, b),
                queryMax(node << 1 | 1, mid + 1, r, a, b));
}
void solve() {
  long long n, m, a, type, y, b, x, k;
  cin >> n >> m;
  for (long long i = 1; i < n + 1; i += 1) {
    cin >> A[i];
    update(1, 1, n, i, A[i]);
  }
  while (m--) {
    cin >> type;
    if (type == 1) {
      cin >> a >> b;
      cout << querySum(1, 1, n, a, b) << "\n";
    } else if (type == 2) {
      cin >> a >> b >> x;
      while (1) {
        long long tmp = queryMax(1, 1, n, a, b);
        if (A[tmp] < x) break;
        y = A[tmp] % x;
        update(1, 1, n, tmp, y);
      }
    } else {
      cin >> k >> x;
      update(1, 1, n, k, x);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (t--) {
    solve();
  }
  return 0;
}