#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 3e5 + 7;
const int M = 1e6 + 7;
int dfn[N], low[N], sccno[N], scc, times;
int mx[N], mn[N], s[N], siz[N];
long long sum[N];
vector<int> G[N];
stack<int> st;
void dfs(int u, int fa) {
  dfn[u] = low[u] = ++times;
  st.push(u);
  for (int v : G[u]) {
    if (v == fa) continue;
    if (!dfn[v]) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
    } else if (!sccno[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (dfn[u] == low[u]) {
    scc++;
    while (1) {
      siz[scc]++;
      int x = st.top();
      st.pop();
      sccno[x] = scc;
      if (x == u) break;
    }
  }
}
signed main() {
  int n, m, q, u, v;
  scanf("%d", &n);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &u);
    scanf("%d", &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++) mx[sccno[i]] = i, s[i] = n;
  for (int i = n; i; i--) mn[sccno[i]] = i;
  for (int i = 1; i <= scc; i++)
    if (siz[i] > 1) s[mn[i]] = mx[i] - 1;
  for (int i = n - 1; i; i--) s[i] = min(s[i], s[i + 1]);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + s[i] - i + 1;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &u);
    scanf("%d", &v);
    int p = lower_bound(s + u, s + v + 1, v) - s;
    long long ans = 0ll;
    ans += sum[p - 1] - sum[u - 1];
    ans += (long long)(v - p + 1ll) * (long long)(v - p + 2ll) / 2ll;
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 3e5 + 7;
const int M = 1e6 + 7;
int dfn[N], low[N], sccno[N], scc, times;
int mx[N], mn[N], s[N], siz[N];
long long sum[N];
vector<int> G[N];
stack<int> st;
void dfs(int u, int fa) {
  dfn[u] = low[u] = ++times;
  st.push(u);
  for (int v : G[u]) {
    if (v == fa) continue;
    if (!dfn[v]) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
    } else if (!sccno[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (dfn[u] == low[u]) {
    scc++;
    while (1) {
      siz[scc]++;
      int x = st.top();
      st.pop();
      sccno[x] = scc;
      if (x == u) break;
    }
  }
}
signed main() {
  int n, m, q, u, v;
  scanf("%d", &n);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &u);
    scanf("%d", &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++) mx[sccno[i]] = i, s[i] = n;
  for (int i = n; i; i--) mn[sccno[i]] = i;
  for (int i = 1; i <= scc; i++)
    if (siz[i] > 1) s[mn[i]] = mx[i] - 1;
  for (int i = n - 1; i; i--) s[i] = min(s[i], s[i + 1]);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + s[i] - i + 1;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &u);
    scanf("%d", &v);
    int p = lower_bound(s + u, s + v + 1, v) - s;
    long long ans = 0ll;
    ans += sum[p - 1] - sum[u - 1];
    ans += (long long)(v - p + 1ll) * (long long)(v - p + 2ll) / 2ll;
    printf("%lld\n", ans);
  }
  return 0;
}

```