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 long long INF = (long long)1e18;
const long long MOD = (long long)1e9 + 7;
long long n, m, ans, x, a[200000], i, j;
set<long long> s[20];
set<long long>::iterator it1, it2;
int main() {
  cin >> n >> m;
  for (j = 0; j < 20; ++j) s[j].insert(0), s[j].insert(n + 1);
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    for (j = 0; j < 20; ++j)
      if (!(a[i] & (1 << j))) s[j].insert(i);
  }
  for (i = 1; i <= n; ++i)
    for (j = 0; j < 20; ++j)
      if ((a[i] & (1 << j))) {
        it1 = s[j].upper_bound(i);
        ans += (*it1 - i) * (1 << j);
      }
  while (m--) {
    cin >> i >> x;
    for (j = 0; j < 20; ++j)
      if ((a[i] & (1 << j)) && !(x & (1 << j))) {
        s[j].insert(i);
        it1 = it2 = s[j].lower_bound(i);
        --it1;
        ++it2;
        ans -= (1 << j) * (i - *it1) * (*it2 - i);
      } else if (!(a[i] & (1 << j)) && (x & (1 << j))) {
        it1 = it2 = s[j].lower_bound(i);
        --it1;
        ++it2;
        ans += (1 << j) * (i - *it1) * (*it2 - i);
        s[j].erase(i);
      }
    a[i] = x;
    cout << ans << endl;
  }
}