#include <vector>
#include <cstdio>
#include <algorithm>
#define FOR(i, l, r) for(int i = l; i <= r; ++i)

using namespace std;

const int inf = 1e9;
const int N = 800010;

int mn[N], b[N];
int n, m, l, r, ans;
vector <int> p[N];

void build(int h, int l, int r)
{
    if (l == r)
    {
        if (!l) mn[h] = 0; else mn[h] = inf;
        return;
    }
    int mid = (l + r) >> 1;
    build(h << 1, l, mid); build(h << 1 | 1, mid + 1, r);
    mn[h] = min(mn[h << 1], mn[h << 1 | 1]);
}

int find(int h, int l, int r, int ll, int rr)
{
    if (l == ll && r == rr) return mn[h];
    int mid = (l + r) >> 1;
    if (rr <= mid) return find(h << 1, l, mid, ll, rr); else
    if (ll > mid) return find(h << 1 | 1, mid + 1, r, ll, rr); else
        return min(find(h << 1, l, mid, ll, mid), find(h << 1 | 1, mid + 1, r, mid + 1, rr));
}

void change(int h, int l, int r, int x, int y)
{
    if (l == r) {mn[h] = min(mn[h], y); return;}
    int mid = (l + r) >> 1;
    if (x <= mid) change(h << 1, l, mid, x, y);
        else change(h << 1 | 1, mid + 1, r, x, y);
    mn[h] = min(mn[h << 1], mn[h << 1 | 1]);
}

int main()
{
    scanf("%d", &n);
    FOR(i, 0, n - 1)
    {
        scanf("%d", &b[i]);
        if (!b[i]) ++ans;
    }
    scanf("%d", &m);
    FOR(i, 1, m)
    {
        scanf("%d%d", &l, &r);
        p[l - 1].push_back(r);
    }
    build(1, 0, n);
    FOR(i, 0, n - 1)
    {
        for(int j = 0; j < p[i].size(); ++j)
            change(1, 0, n, p[i][j], find(1, 0, n, i, p[i][j]));
        change(1, 0, n, i + 1, find(1, 0, n, i, i) + (b[i] ? 1 : -1));
    }
    printf("%d\n", find(1, 0, n, n, n) + ans);
    return 0;
}

### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <vector>
#include <cstdio>
#include <algorithm>
#define FOR(i, l, r) for(int i = l; i <= r; ++i)

using namespace std;

const int inf = 1e9;
const int N = 800010;

int mn[N], b[N];
int n, m, l, r, ans;
vector <int> p[N];

void build(int h, int l, int r)
{
    if (l == r)
    {
        if (!l) mn[h] = 0; else mn[h] = inf;
        return;
    }
    int mid = (l + r) >> 1;
    build(h << 1, l, mid); build(h << 1 | 1, mid + 1, r);
    mn[h] = min(mn[h << 1], mn[h << 1 | 1]);
}

int find(int h, int l, int r, int ll, int rr)
{
    if (l == ll && r == rr) return mn[h];
    int mid = (l + r) >> 1;
    if (rr <= mid) return find(h << 1, l, mid, ll, rr); else
    if (ll > mid) return find(h << 1 | 1, mid + 1, r, ll, rr); else
        return min(find(h << 1, l, mid, ll, mid), find(h << 1 | 1, mid + 1, r, mid + 1, rr));
}

void change(int h, int l, int r, int x, int y)
{
    if (l == r) {mn[h] = min(mn[h], y); return;}
    int mid = (l + r) >> 1;
    if (x <= mid) change(h << 1, l, mid, x, y);
        else change(h << 1 | 1, mid + 1, r, x, y);
    mn[h] = min(mn[h << 1], mn[h << 1 | 1]);
}

int main()
{
    scanf("%d", &n);
    FOR(i, 0, n - 1)
    {
        scanf("%d", &b[i]);
        if (!b[i]) ++ans;
    }
    scanf("%d", &m);
    FOR(i, 1, m)
    {
        scanf("%d%d", &l, &r);
        p[l - 1].push_back(r);
    }
    build(1, 0, n);
    FOR(i, 0, n - 1)
    {
        for(int j = 0; j < p[i].size(); ++j)
            change(1, 0, n, p[i][j], find(1, 0, n, i, p[i][j]));
        change(1, 0, n, i + 1, find(1, 0, n, i, i) + (b[i] ? 1 : -1));
    }
    printf("%d\n", find(1, 0, n, n, n) + ans);
    return 0;
}

```