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;
void dout() { cerr << endl; }
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
  cerr << H << ' ';
  dout(T...);
}
using ll = long long;
using pii = pair<ll, ll>;
using ld = long double;
template <typename T>
void do_uniq(vector<T> &vec) {
  sort((vec).begin(), (vec).end());
  vec.resize(unique((vec).begin(), (vec).end()) - vec.begin());
}
int mod = 1e9 + 7;
int ml(int a, int b) { return (a * 1ll * b) % mod; }
int add(int a, int b) { return (a + b) % mod; }
int sub(int a, int b) { return add(a, mod - b); }
int sq(int a) { return ml(a, a); }
int b_p(int b, int p) {
  if (p == 0) return 1;
  if (p & 1) return ml(b, b_p(b, p - 1));
  return sq(b_p(b, p >> 1));
}
ld INF = 1000;
const int MAXN = 1e5 + 12;
ll mx[8 * MAXN];
ll sm[8 * MAXN];
void update(int v, int l, int r, int tl, int tr, int mod) {
  if (tl > tr) return;
  if (mx[v] < mod) return;
  if (l == r) {
    sm[v] %= mod;
    mx[v] %= mod;
    return;
  }
  int mid = (l + r) >> 1;
  update(2 * v, l, mid, tl, min(mid, tr), mod);
  update(1 + 2 * v, mid + 1, r, max(mid + 1, tl), tr, mod);
  mx[v] = max(mx[2 * v], mx[2 * v + 1]);
  sm[v] = sm[2 * v] + sm[2 * v + 1];
}
ll sum(int v, int l, int r, int tl, int tr) {
  if (tl > tr) return 0;
  if (tl == l && tr == r) return sm[v];
  int mid = (l + r) >> 1;
  return sum(2 * v, l, mid, tl, min(mid, tr)) +
         sum(1 + 2 * v, mid + 1, r, max(mid + 1, tl), tr);
}
void sett(int v, int l, int r, int pos, int x) {
  if (l == r) {
    mx[v] = sm[v] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    sett(v * 2, l, mid, pos, x);
  else
    sett(1 + 2 * v, mid + 1, r, pos, x);
  mx[v] = max(mx[2 * v], mx[2 * v + 1]);
  sm[v] = sm[2 * v] + sm[2 * v + 1];
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    int curr;
    cin >> curr;
    sett(1, 0, n - 1, i, curr);
  }
  for (int i = 0; i < m; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      --l;
      --r;
      cout << sum(1, 0, n - 1, l, r) << "\n";
    } else if (t == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      --l;
      --r;
      update(1, 0, n - 1, l, r, x);
    } else {
      int k, x;
      cin >> k >> x;
      --k;
      sett(1, 0, n - 1, k, x);
    }
  }
}
void multisolve() {
  int t;
  cin >> t;
  int i = 1;
  while (t--) {
    solve();
    i++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cout << fixed << setprecision(20);
  solve();
}