#include <bits/stdc++.h>
using namespace std;
class SegmentTree {
  vector<int> stmin, stmax;
  int n;

 public:
  SegmentTree(int sz) {
    for (n = 1; n < sz; n <<= 1)
      ;
    stmin.assign(n << 1, 300009);
    stmax.assign(n << 1, 0);
  }
  void update(int i, int x) {
    stmax[i + n] = x;
    stmin[i + n] = x;
    for (i += n, i >>= 1; i; i >>= 1) {
      stmax[i] = max(stmax[i << 1], stmax[1 + (i << 1)]);
      stmin[i] = min(stmin[i << 1], stmin[1 + (i << 1)]);
    }
  }
  int maxquery(int l, int r) {
    int ans = 0;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = max(ans, stmax[l++]);
      if (r & 1) ans = max(ans, stmax[--r]);
    }
    return ans;
  }
  int minquery(int l, int r) {
    int ans = 300009;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = min(ans, stmin[l++]);
      if (r & 1) ans = min(ans, stmin[--r]);
    }
    return ans;
  }
};
SegmentTree st(300009);
int num[300009], pos[300009], last[300009], am[300009], cnt, N, M;
long long sum[300009];
vector<int> adjList[300009];
void dfs(int u, int p) {
  num[u] = 1;
  pos[u] = cnt++;
  st.update(pos[u], u);
  for (int i = 0; i < (int)adjList[u].size(); i++) {
    int v = adjList[u][i];
    if (v == p) continue;
    if (num[v] == 0)
      dfs(v, u);
    else if (num[v] == 1) {
      int mi = st.minquery(pos[v], pos[u]);
      int ma = st.maxquery(pos[v], pos[u]);
      last[mi] = min(last[mi], ma - 1);
    }
  }
  num[u] = 2;
  cnt--;
}
int main() {
  scanf("%d %d", &N, &M);
  for (int j = 0; j < M; j++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  for (int u = 1; u <= N; u++) {
    last[u] = N;
    num[u] = 0;
  }
  cnt = 0;
  for (int u = 1; u <= N; u++) {
    if (num[u] == 0) dfs(u, -1);
  }
  for (int u = N - 1; u >= 1; u--) {
    last[u] = min(last[u], last[u + 1]);
  }
  sum[0] = 0;
  for (int u = 1; u <= N; u++) {
    am[u] = last[u] - u + 1;
    sum[u] = am[u] + sum[u - 1];
  }
  int Q, l, r;
  scanf("%d", &Q);
  for (int q = 0; q < Q; q++) {
    scanf("%d %d", &l, &r);
    int lo = l - 1;
    int hi = r + 1;
    while (hi > lo + 1) {
      int mid = (hi + lo) / 2;
      if (last[mid] > r)
        hi = mid;
      else
        lo = mid;
    }
    int n = r - hi + 1;
    long long ans = n * (n + 1LL) / 2 + sum[lo] - sum[l - 1];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
class SegmentTree {
  vector<int> stmin, stmax;
  int n;

 public:
  SegmentTree(int sz) {
    for (n = 1; n < sz; n <<= 1)
      ;
    stmin.assign(n << 1, 300009);
    stmax.assign(n << 1, 0);
  }
  void update(int i, int x) {
    stmax[i + n] = x;
    stmin[i + n] = x;
    for (i += n, i >>= 1; i; i >>= 1) {
      stmax[i] = max(stmax[i << 1], stmax[1 + (i << 1)]);
      stmin[i] = min(stmin[i << 1], stmin[1 + (i << 1)]);
    }
  }
  int maxquery(int l, int r) {
    int ans = 0;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = max(ans, stmax[l++]);
      if (r & 1) ans = max(ans, stmax[--r]);
    }
    return ans;
  }
  int minquery(int l, int r) {
    int ans = 300009;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = min(ans, stmin[l++]);
      if (r & 1) ans = min(ans, stmin[--r]);
    }
    return ans;
  }
};
SegmentTree st(300009);
int num[300009], pos[300009], last[300009], am[300009], cnt, N, M;
long long sum[300009];
vector<int> adjList[300009];
void dfs(int u, int p) {
  num[u] = 1;
  pos[u] = cnt++;
  st.update(pos[u], u);
  for (int i = 0; i < (int)adjList[u].size(); i++) {
    int v = adjList[u][i];
    if (v == p) continue;
    if (num[v] == 0)
      dfs(v, u);
    else if (num[v] == 1) {
      int mi = st.minquery(pos[v], pos[u]);
      int ma = st.maxquery(pos[v], pos[u]);
      last[mi] = min(last[mi], ma - 1);
    }
  }
  num[u] = 2;
  cnt--;
}
int main() {
  scanf("%d %d", &N, &M);
  for (int j = 0; j < M; j++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  for (int u = 1; u <= N; u++) {
    last[u] = N;
    num[u] = 0;
  }
  cnt = 0;
  for (int u = 1; u <= N; u++) {
    if (num[u] == 0) dfs(u, -1);
  }
  for (int u = N - 1; u >= 1; u--) {
    last[u] = min(last[u], last[u + 1]);
  }
  sum[0] = 0;
  for (int u = 1; u <= N; u++) {
    am[u] = last[u] - u + 1;
    sum[u] = am[u] + sum[u - 1];
  }
  int Q, l, r;
  scanf("%d", &Q);
  for (int q = 0; q < Q; q++) {
    scanf("%d %d", &l, &r);
    int lo = l - 1;
    int hi = r + 1;
    while (hi > lo + 1) {
      int mid = (hi + lo) / 2;
      if (last[mid] > r)
        hi = mid;
      else
        lo = mid;
    }
    int n = r - hi + 1;
    long long ans = n * (n + 1LL) / 2 + sum[lo] - sum[l - 1];
    printf("%I64d\n", ans);
  }
  return 0;
}

```