Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
vector<int> t;
long long s = 0;
void find(int k, long long& l, long long& r, int x, int n) {
  l = r = 0;
  for (auto i = x - 1; i; i--) {
    if ((t[i] & k) == 0) break;
    l++;
  }
  for (auto i = x + 1; i <= n; i++) {
    if ((t[i] & k) == 0) break;
    r++;
  }
}
void solve(int p, int x, int y, int n) {
  long long l, r;
  for (auto i = 0; (1 << i) < N; i++) {
    auto k = (1 << i);
    if ((x & k) == (y & k)) continue;
    find(k, l, r, p, n);
    auto u = (l * r + l + r + 1) * (long long)k;
    if (x & k) {
      s += u;
    } else {
      s -= u;
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  t.resize(n + 1);
  for (auto i = 1; i <= n; i++) cin >> t[i];
  for (auto i = 0; (1 << i) < N; i++) {
    auto k = (1 << i);
    long long c = 0;
    for (auto j = 1; j <= n; j++) {
      if (t[j] & k)
        c++;
      else {
        if (c) s += (c * (c + 1) * (long long)k / 2);
        c = 0;
      }
    }
    if (c) s += (c * (c + 1) * (long long)k / 2);
  }
  int p, x;
  for (auto i = 0; i < m; i++) {
    cin >> p >> x;
    solve(p, x, t[p], n);
    t[p] = x;
    cout << s << endl;
  }
  return 0;
}