#include <iostream>
#include <vector>
#include <algorithm>

#ifndef INT_MAX
#define INT_MAX 2147483647
#endif

#define rep(i,i0,n) for (int (i) = (i0); (i) < (n); ++(i))

using namespace std;

const int MAX_N = 262144;

class SegTree {
public:
    SegTree();
    void update(int i, int x);
    int query(int a, int b, int k, int l, int r);
private:
    int dat[MAX_N * 2];
    int n;
};

SegTree::SegTree() {
    n = MAX_N;
    rep(i, 0, 2 * n) {
        dat[i] = INT_MAX;
    }
}

void SegTree::update(int i, int x) {
    i += n - 1;
    dat[i] = x;
    while (0 < i) {
        i = (i - 1) / 2;
        dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
}

int SegTree::query(int a, int b, int k = 0, int l = 0, int r = MAX_N) {
    if (r <= a || b <= l) return INT_MAX;
    if (a <= l && r <= b) return dat[k];
    else {
        int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return min(vl, vr);
    }
}

int N = 0;
int Q = 0;
vector<int> L[202020];
int B[202020];
int cnt = 0;

int solve() {
    SegTree seg;
    int d = 0;

    rep(i, 0, N) {
        for(int r : L[i]) {
            seg.update(r, min(d, seg.query(i, r + 1)));
        }
        d = min(d + (B[i] ? 1 : -1), seg.query(i, i + 1));
    }

    return d + cnt;
}

int main(int argc, const char * argv[]) {
    int b, l, r;

    cin >> N;
    rep(i, 0, N) {
        cin >> b;
        B[i] = b;
        if (!b) ++cnt;
    }

    cin >> Q;
    rep(i, 0, Q) {
        cin >> l >> r;
        --l;
        --r;
        L[l].push_back(r);
    }

    cout << solve() << endl;

    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <iostream>
#include <vector>
#include <algorithm>

#ifndef INT_MAX
#define INT_MAX 2147483647
#endif

#define rep(i,i0,n) for (int (i) = (i0); (i) < (n); ++(i))

using namespace std;

const int MAX_N = 262144;

class SegTree {
public:
    SegTree();
    void update(int i, int x);
    int query(int a, int b, int k, int l, int r);
private:
    int dat[MAX_N * 2];
    int n;
};

SegTree::SegTree() {
    n = MAX_N;
    rep(i, 0, 2 * n) {
        dat[i] = INT_MAX;
    }
}

void SegTree::update(int i, int x) {
    i += n - 1;
    dat[i] = x;
    while (0 < i) {
        i = (i - 1) / 2;
        dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
}

int SegTree::query(int a, int b, int k = 0, int l = 0, int r = MAX_N) {
    if (r <= a || b <= l) return INT_MAX;
    if (a <= l && r <= b) return dat[k];
    else {
        int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return min(vl, vr);
    }
}

int N = 0;
int Q = 0;
vector<int> L[202020];
int B[202020];
int cnt = 0;

int solve() {
    SegTree seg;
    int d = 0;

    rep(i, 0, N) {
        for(int r : L[i]) {
            seg.update(r, min(d, seg.query(i, r + 1)));
        }
        d = min(d + (B[i] ? 1 : -1), seg.query(i, i + 1));
    }

    return d + cnt;
}

int main(int argc, const char * argv[]) {
    int b, l, r;

    cin >> N;
    rep(i, 0, N) {
        cin >> b;
        B[i] = b;
        if (!b) ++cnt;
    }

    cin >> Q;
    rep(i, 0, Q) {
        cin >> l >> r;
        --l;
        --r;
        L[l].push_back(r);
    }

    cout << solve() << endl;

    return 0;
}

```