Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int n_ = 200000+2;
const int sz_ = (1<<18);

int n, q, a[n_], s[n_], z[n_];
int d[n_];
int tree[sz_+sz_];
vector<int> v[n_];

void update(int s, int e, int x) {
    s += sz_, e += sz_;
    while (s <= e) {
        tree[s] = min(tree[s], x);
        tree[e] = min(tree[e], x);
        s  = (s+1)>>1, e = (e-1)>>1;
    }
}

int query(int i) {
    int r = 1e9;
    i += sz_;
    while (i) {
        r = min(r, tree[i]);
        i >>= 1;
    }
    return r;
}

int main() {
    for (int i = 1; i < sz_+sz_; i++) {
        tree[i] = 1e9;
    }
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        s[i] = s[i-1] + 1 - a[i];
    }
    scanf("%d", &q);
    while (q--) {
        int l, r;
        scanf("%d %d", &l, &r);
        v[r].push_back(l);
    }

    for (int i = 1; i <= n; i++) {
        update(i-1, i-1, d[i-1]-s[i-1]);
        d[i] = d[i-1] + a[i];
        for (int l : v[i]) {
            update(l, i, query(l-1));
        }
        d[i] = min(d[i], query(i)+s[i]);
    }

    //puts("");

    printf("%d\n", d[n]);

    return 0;
}