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;
unsigned long long n, m, a[1000001], ans[1000001 << 2], tag[1000001 << 2];
unsigned long long Mx[1000001 << 2];
template <class T>
inline bool read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
inline long long ls(long long x) { return x << 1; }
inline long long rs(long long x) { return x << 1 | 1; }
inline void push_up(long long p) {
  ans[p] = ans[ls(p)] + ans[rs(p)];
  Mx[p] = max(Mx[ls(p)], Mx[rs(p)]);
}
void build(long long p, long long l, long long r) {
  if (l == r) {
    ans[p] = a[l];
    Mx[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(ls(p), l, mid);
  build(rs(p), mid + 1, r);
  push_up(p);
}
inline void update(long long nl, long long nr, long long l, long long r,
                   long long p, long long k) {
  if (l == r) {
    ans[p] = k;
    Mx[p] = k;
    return;
  }
  long long mid = (l + r) >> 1;
  if (nl <= mid) update(nl, nr, l, mid, ls(p), k);
  if (nr > mid) update(nl, nr, mid + 1, r, rs(p), k);
  push_up(p);
}
inline void updata1(long long nl, long long nr, long long l, long long r,
                    long long p, long long Mod) {
  if (Mx[p] < Mod) return;
  if (l == r && Mod <= ans[p]) {
    ans[p] = ans[p] % Mod;
    Mx[p] = ans[p];
    return;
  }
  long long mid = (l + r) >> 1;
  if (nl <= mid) updata1(nl, nr, l, mid, ls(p), Mod);
  if (nr > mid) updata1(nl, nr, mid + 1, r, rs(p), Mod);
  push_up(p);
}
long long query(long long q_x, long long q_y, long long l, long long r,
                long long p) {
  long long res = 0;
  if (q_x <= l && r <= q_y) return ans[p];
  long long mid = (l + r) >> 1;
  if (q_x <= mid) res += query(q_x, q_y, l, mid, ls(p));
  if (q_y > mid) res += query(q_x, q_y, mid + 1, r, rs(p));
  return res;
}
int main() {
  long long a1, b, c, d, e, f;
  read(n);
  read(m);
  for (int i = (1); i <= (n); i++) read(a[i]);
  build(1, 1, n);
  while (m--) {
    scanf("%lld", &a1);
    switch (a1) {
      case 3: {
        scanf("%lld%lld", &b, &c);
        update(b, b, 1, n, 1, c);
        break;
      }
      case 1: {
        scanf("%lld%lld", &e, &f);
        printf("%lld\n", query(e, f, 1, n, 1));
        break;
      }
      case 2: {
        scanf("%lld%lld%lld", &b, &c, &d);
        updata1(b, c, 1, n, 1, d);
      }
    }
  }
  return 0;
}