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 + 19, Bit = 30;
int A[Maxn];
int n, m, x, v;
long long Ans;
set<int> S[Bit];
inline int is_1(int a, int Dis) { return (a >> Dis) & 1; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 0; i < Bit; i++)
    for (int j = 0; j <= n + 1; j++) {
      int tmp = is_1(A[j], i);
      if (!is_1(A[j], i)) S[i].insert(j);
    }
  for (int i = 0; i < Bit; i++)
    if (S[i].size() > 1)
      for (set<int>::iterator j = ++S[i].begin(); j != S[i].end(); j++) {
        set<int>::iterator k = j;
        --k;
        Ans += (1LL << i) * (*j - *k - 1) * (*j - *k) >> 1;
      }
  while (m--) {
    scanf("%d%d", &x, &v);
    for (int j = 0; j < Bit; j++) {
      set<int>::iterator L = --S[j].lower_bound(x);
      set<int>::iterator R = S[j].upper_bound(x);
      long long delta =
          (1LL * (*R - *L - 1) * (*R - *L) >> 1) -
          ((1LL * (x - *L - 1) * (x - *L) + 1LL * (*R - x - 1) * (*R - x)) >>
           1);
      if (is_1(A[x], j) && !is_1(v, j)) {
        Ans -= delta * (1LL << j);
        S[j].insert(x);
      }
      if (!is_1(A[x], j) && is_1(v, j)) {
        Ans += delta * (1LL << j);
        S[j].erase(x);
      }
    }
    A[x] = v;
    printf("%I64d\n", Ans);
  }
}