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 MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e5 + 10;
const long long INF = 1e18;
struct node {
  long long mx = -INF, mn = INF;
  bool lz = 0;
  long long sum = 0;
  long long len = 0;
} st[MAXN << 2];
node merge(node a, node b) {
  node c;
  c.mx = max(a.mx, b.mx);
  c.mn = min(a.mn, b.mn);
  c.sum = a.sum + b.sum;
  c.len = a.len + b.len;
  return c;
}
void weed(int v) {
  if (!st[v].lz) return;
  assert(st[v].mx == st[v].mn);
  for (int u : {v << 1, v << 1 ^ 1}) {
    st[u].sum = st[v].mx * st[u].len;
    st[u].mx = st[u].mn = st[v].mx;
    st[u].lz = 1;
  }
  st[v].lz = 0;
}
long long n, a[MAXN];
void plant(int v, int b, int e) {
  if (e - b == 1) {
    st[v].mx = st[v].mn = st[v].sum = a[b];
    st[v].len = 1;
    return;
  }
  int m = (b + e) >> 1;
  plant(v << 1, b, m);
  plant(v << 1 ^ 1, m, e);
  st[v] = merge(st[v << 1], st[v << 1 ^ 1]);
}
long long smoke(int v, int b, int e, int l, int r) {
  if (l >= r) return 0;
  if (b == l && e == r) {
    return st[v].sum;
  }
  weed(v);
  int m = (b + e) >> 1;
  return smoke(v << 1, b, m, l, min(r, m)) +
         smoke(v << 1 ^ 1, m, e, max(l, m), r);
}
void water(int v, int b, int e, int l, int r, int mod) {
  if (l >= r) return;
  if (st[v].mx < mod) return;
  if (b == l && e == r && st[v].mx == st[v].mn) {
    st[v].mx = st[v].mn = st[v].mx % mod;
    st[v].sum = st[v].len * st[v].mx;
    st[v].lz = 1;
    return;
  }
  weed(v);
  int m = (b + e) >> 1;
  water(v << 1, b, m, l, min(r, m), mod);
  water(v << 1 ^ 1, m, e, max(l, m), r, mod);
  st[v] = merge(st[v << 1], st[v << 1 ^ 1]);
}
void nut(int v, int b, int e, int i, int x) {
  if (e - b == 1) {
    st[v].mx = st[v].mn = st[v].sum = x;
    st[v].len = 1;
    return;
  }
  weed(v);
  int m = (b + e) >> 1;
  if (i < m)
    nut(v << 1, b, m, i, x);
  else
    nut(v << 1 ^ 1, m, e, i, x);
  st[v] = merge(st[v << 1], st[v << 1 ^ 1]);
}
void solve() {
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  plant(1, 0, n);
  while (q--) {
    int qt;
    cin >> qt;
    if (qt == 1) {
      int l, r;
      cin >> l >> r;
      --l;
      cout << smoke(1, 0, n, l, r) << "\n";
    } else if (qt == 2) {
      int l, r, mod;
      cin >> l >> r >> mod;
      --l;
      water(1, 0, n, l, r, mod);
    } else {
      int i, x;
      cin >> i >> x;
      --i;
      nut(1, 0, n, i, x);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}