#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1500000001;
const int maxn = 3e5 + 7;
const double pi = acos(-1.0);
const int v2 = (mod + 1) / 2;
const double eps = 1e-7;
int n, m, l[maxn], r[maxn], cnt, maxl[maxn], pos[maxn], dep[maxn], vis[maxn],
    father[maxn];
long long sum[maxn];
vector<int> g[maxn];
void dfs(int now, int fa, int num) {
  vis[now] = 1, father[now] = fa, dep[now] = num;
  for (int i = 0; i < g[now].size(); i++) {
    int to = g[now][i];
    if (!vis[to])
      dfs(to, now, num + 1);
    else if (to != fa && dep[to] < dep[now]) {
      cnt++;
      for (int j = now;; j = father[j]) {
        l[cnt] = min(l[cnt], j), r[cnt] = max(r[cnt], j);
        if (j == to) break;
      }
    }
  }
}
void init() {
  memset(maxl, 0, sizeof(maxl));
  for (int i = 0; i <= cnt; i++) {
    maxl[r[i]] = max(maxl[r[i]], l[i]);
  }
  for (int i = 1; i <= n; i++) {
    maxl[i] = max(maxl[i], maxl[i - 1]);
  }
  for (int i = n; i > maxl[n]; i--) pos[i] = n + 1;
  for (int i = 1; i <= n; i++) {
    for (int j = maxl[i]; j > maxl[i - 1]; j--) pos[j] = i;
  }
  sum[0] = 0;
  for (int i = 0; i <= n; i++) {
    sum[i] = sum[i - 1] + 1ll * (i - maxl[i]);
  }
}
long long solve(int a, int b) {
  int p = pos[--a];
  if (p <= a) {
    return sum[b] - sum[a];
  } else if (p > a && p <= b)
    return sum[b] - sum[p - 1] + 1ll * (p - a - 1) * (p - a) / 2;
  else
    return 1ll * (b - a + 1) * (b - a) / 2;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i <= n; i++) vis[i] = 0, l[i] = n + 1, r[i] = 0;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  cnt = -1;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, -1, 0);
  }
  init();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%lld\n", solve(a, b));
  }
  return 0;
}

### Prompt

Generate 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;
const int INF = 0x3f3f3f3f;
const int mod = 1500000001;
const int maxn = 3e5 + 7;
const double pi = acos(-1.0);
const int v2 = (mod + 1) / 2;
const double eps = 1e-7;
int n, m, l[maxn], r[maxn], cnt, maxl[maxn], pos[maxn], dep[maxn], vis[maxn],
    father[maxn];
long long sum[maxn];
vector<int> g[maxn];
void dfs(int now, int fa, int num) {
  vis[now] = 1, father[now] = fa, dep[now] = num;
  for (int i = 0; i < g[now].size(); i++) {
    int to = g[now][i];
    if (!vis[to])
      dfs(to, now, num + 1);
    else if (to != fa && dep[to] < dep[now]) {
      cnt++;
      for (int j = now;; j = father[j]) {
        l[cnt] = min(l[cnt], j), r[cnt] = max(r[cnt], j);
        if (j == to) break;
      }
    }
  }
}
void init() {
  memset(maxl, 0, sizeof(maxl));
  for (int i = 0; i <= cnt; i++) {
    maxl[r[i]] = max(maxl[r[i]], l[i]);
  }
  for (int i = 1; i <= n; i++) {
    maxl[i] = max(maxl[i], maxl[i - 1]);
  }
  for (int i = n; i > maxl[n]; i--) pos[i] = n + 1;
  for (int i = 1; i <= n; i++) {
    for (int j = maxl[i]; j > maxl[i - 1]; j--) pos[j] = i;
  }
  sum[0] = 0;
  for (int i = 0; i <= n; i++) {
    sum[i] = sum[i - 1] + 1ll * (i - maxl[i]);
  }
}
long long solve(int a, int b) {
  int p = pos[--a];
  if (p <= a) {
    return sum[b] - sum[a];
  } else if (p > a && p <= b)
    return sum[b] - sum[p - 1] + 1ll * (p - a - 1) * (p - a) / 2;
  else
    return 1ll * (b - a + 1) * (b - a) / 2;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i <= n; i++) vis[i] = 0, l[i] = n + 1, r[i] = 0;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  cnt = -1;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, -1, 0);
  }
  init();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%lld\n", solve(a, b));
  }
  return 0;
}

```