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>
const double pi = acos(-1);
template <typename _t>
inline void read(_t &x) {
  x = 0;
  _t fu = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fu = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch & 15);
    ch = getchar();
  }
  x *= fu;
}
long long gcd(long long x, long long y) { return !x ? y : gcd(y % x, x); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
struct Mart {
  int a, b, c, d;
  Mart operator*(Mart &m) {
    Mart temp;
    temp.a = a * m.a + b * m.c;
    temp.b = a * m.b + b * m.d;
    temp.c = c * m.a + d * m.c;
    temp.d = c * m.b + d * m.d;
    return temp;
  }
};
const int maxn = 1e5 + 10;
int a[maxn];
long long st[maxn << 2], ms[maxn << 2];
void up(int p) {
  st[p] = st[(p << 1)] + st[(p << 1 | 1)];
  ms[p] = std::max(ms[(p << 1)], ms[(p << 1 | 1)]);
}
void build(int p, int l, int r) {
  if (l == r) {
    st[p] = a[l];
    ms[p] = a[l];
    return;
  }
  build((p << 1), l, ((l + r) >> 1));
  build((p << 1 | 1), ((l + r) >> 1) + 1, r);
  up(p);
}
void update2(int p, int l, int r, int L, int R, int m) {
  if (l == r) {
    st[p] %= m;
    ms[p] %= m;
    return;
  }
  if (L <= l && r <= R) {
    if (ms[p] >= m) {
      if (ms[(p << 1)] >= m) update2((p << 1), l, ((l + r) >> 1), L, R, m);
      if (ms[(p << 1 | 1)] >= m)
        update2((p << 1 | 1), ((l + r) >> 1) + 1, r, L, R, m);
      up(p);
    }
    return;
  }
  if (L <= ((l + r) >> 1) && ms[(p << 1)] >= m)
    update2((p << 1), l, ((l + r) >> 1), L, R, m);
  if (R > ((l + r) >> 1) && ms[(p << 1 | 1)] >= m)
    update2((p << 1 | 1), ((l + r) >> 1) + 1, r, L, R, m);
  up(p);
}
void update3(int p, int l, int r, int x, int k) {
  if (l == r && l == x) {
    st[p] = k;
    ms[p] = k;
    return;
  }
  if (x <= ((l + r) >> 1))
    update3((p << 1), l, ((l + r) >> 1), x, k);
  else
    update3((p << 1 | 1), ((l + r) >> 1) + 1, r, x, k);
  up(p);
}
long long query(int p, int l, int r, int L, int R) {
  if (L <= l && r <= R) return st[p];
  long long ans = 0;
  if (L <= ((l + r) >> 1)) ans += query((p << 1), l, ((l + r) >> 1), L, R);
  if (R > ((l + r) >> 1))
    ans += query((p << 1 | 1), ((l + r) >> 1) + 1, r, L, R);
  return ans;
}
int main() {
  int n, m;
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  build(1, 1, n);
  int x, y, M;
  for (int i = 0; i < m; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      read(x);
      read(y);
      printf("%lld\n", query(1, 1, n, x, y));
    } else if (op == 2) {
      read(x);
      read(y);
      read(M);
      update2(1, 1, n, x, y, M);
    } else {
      read(x);
      read(y);
      update3(1, 1, n, x, y);
    }
  }
  return 0;
}