// ARC085F.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 2e5 + 200, INF = 0x3f3f3f3f;

#define lson (p << 1)
#define rson ((p << 1) | 1)
#define mid ((l + r) >> 1)

int n, q, nodes[MAX_N << 2], dp[MAX_N], bi[MAX_N];
vector<int> pset[MAX_N];

void build(int l, int r, int p)
{
    if (l == r)
        return (void)(nodes[p] = INF);
    build(l, mid, lson), build(mid + 1, r, rson);
    nodes[p] = min(nodes[lson], nodes[rson]);
}

int query(int ql, int qr, int l, int r, int p)
{
    if (ql <= l && r <= qr)
        return nodes[p];
    int ret = INF;
    if (ql <= mid)
        ret = min(ret, query(ql, qr, l, mid, lson));
    if (mid < qr)
        ret = min(ret, query(ql, qr, mid + 1, r, rson));
    return ret;
}

void update(int qx, int l, int r, int p, int val)
{
    if (l == r)
        return (void)(nodes[p] = min(nodes[p], val));
    if (qx <= mid)
        update(qx, l, mid, lson, val);
    else
        update(qx, mid + 1, r, rson, val);
    nodes[p] = min(nodes[lson], nodes[rson]);
}

int main()
{
    int zeros = 0;
    scanf("%d", &n), build(1, n, 1);
    for (int i = 1; i <= n; i++)
        scanf("%d", &bi[i]), zeros += (1 - bi[i]);
    scanf("%d", &q);
    while (q--)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        pset[l].push_back(r);
    }
    memset(dp, 0x3f, sizeof(dp)), dp[0] = 0;
    for (int l = 1; l <= n; l++)
    {
        for (int r : pset[l])
        {
            int ret = dp[l - 1];
            ret = min(ret, query(max(l - 1, 1), r, 1, n, 1));
            if (ret < dp[r])
                dp[r] = ret, update(r, 1, n, 1, ret);
        }
        dp[l] = min(dp[l], dp[l - 1] + (bi[l] == 1 ? 1 : -1));
    }
    printf("%d\n", dp[n] + zeros);
    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
// ARC085F.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 2e5 + 200, INF = 0x3f3f3f3f;

#define lson (p << 1)
#define rson ((p << 1) | 1)
#define mid ((l + r) >> 1)

int n, q, nodes[MAX_N << 2], dp[MAX_N], bi[MAX_N];
vector<int> pset[MAX_N];

void build(int l, int r, int p)
{
    if (l == r)
        return (void)(nodes[p] = INF);
    build(l, mid, lson), build(mid + 1, r, rson);
    nodes[p] = min(nodes[lson], nodes[rson]);
}

int query(int ql, int qr, int l, int r, int p)
{
    if (ql <= l && r <= qr)
        return nodes[p];
    int ret = INF;
    if (ql <= mid)
        ret = min(ret, query(ql, qr, l, mid, lson));
    if (mid < qr)
        ret = min(ret, query(ql, qr, mid + 1, r, rson));
    return ret;
}

void update(int qx, int l, int r, int p, int val)
{
    if (l == r)
        return (void)(nodes[p] = min(nodes[p], val));
    if (qx <= mid)
        update(qx, l, mid, lson, val);
    else
        update(qx, mid + 1, r, rson, val);
    nodes[p] = min(nodes[lson], nodes[rson]);
}

int main()
{
    int zeros = 0;
    scanf("%d", &n), build(1, n, 1);
    for (int i = 1; i <= n; i++)
        scanf("%d", &bi[i]), zeros += (1 - bi[i]);
    scanf("%d", &q);
    while (q--)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        pset[l].push_back(r);
    }
    memset(dp, 0x3f, sizeof(dp)), dp[0] = 0;
    for (int l = 1; l <= n; l++)
    {
        for (int r : pset[l])
        {
            int ret = dp[l - 1];
            ret = min(ret, query(max(l - 1, 1), r, 1, n, 1));
            if (ret < dp[r])
                dp[r] = ret, update(r, 1, n, 1, ret);
        }
        dp[l] = min(dp[l], dp[l - 1] + (bi[l] == 1 ? 1 : -1));
    }
    printf("%d\n", dp[n] + zeros);
    return 0;
}
```