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 MAXN = 100000 + 10, L = 18;
int n, m, a[MAXN];
set<int> s[L];
int main() {
  cin >> n >> m;
  for (int i = 0; i < L; ++i) s[i].insert(0), s[i].insert(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    for (int j = 0; j < L; ++j)
      if (!(a[i] & (1 << j))) s[j].insert(i);
  }
  long long Tohka = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < L; ++j)
      if (a[i] & (1 << j)) {
        set<int>::iterator it = s[j].lower_bound(i);
        Tohka += (1ll << j) * (*it - i);
      }
  for (int tt = 1; tt <= m; ++tt) {
    int p, first;
    cin >> p >> first;
    for (int j = 0; j < L; ++j)
      if (!(a[p] & (1 << j)) && (first & (1 << j))) {
        set<int>::iterator it = s[j].lower_bound(p);
        ++it;
        int r = *it;
        --it, --it;
        int l = *it;
        Tohka += (1ll << j) * (p - l) * (r - p);
        s[j].erase(p);
      } else if ((a[p] & (1 << j)) && !(first & (1 << j))) {
        s[j].insert(p);
        set<int>::iterator it = s[j].lower_bound(p);
        ++it;
        int r = *it;
        --it, --it;
        int l = *it;
        Tohka -= (1ll << j) * (p - l) * (r - p);
      }
    cout << Tohka << endl;
    a[p] = first;
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}