Problem: The company X has n employees numbered from 1 through n. Each employee u has a direct boss p_u (1 ≤ p_u ≤ n), except for the employee 1 who has no boss. It is guaranteed, that values p_i form a tree. Employee u is said to be in charge of employee v if u is the direct boss of v or there is an employee w such that w is in charge of v and u is the direct boss of w. Also, any employee is considered to be in charge of himself.

In addition, for each employee u we define it's level lv(u) as follow: 

  * lv(1)=0 
  * lv(u)=lv(p_u)+1 for u ≠ 1 



In the near future, there are q possible plans for the company to operate. The i-th plan consists of two integers l_i and r_i, meaning that all the employees in the range [l_i, r_i], and only they, are involved in this plan. To operate the plan smoothly, there must be a project manager who is an employee in charge of all the involved employees. To be precise, if an employee u is chosen as the project manager for the i-th plan then for every employee v ∈ [l_i, r_i], u must be in charge of v. Note, that u is not necessary in the range [l_i, r_i]. Also, u is always chosen in such a way that lv(u) is as large as possible (the higher the level is, the lower the salary that the company has to pay the employee).

Before any plan is operated, the company has JATC take a look at their plans. After a glance, he tells the company that for every plan, it's possible to reduce the number of the involved employees exactly by one without affecting the plan. Being greedy, the company asks JATC which employee they should kick out of the plan so that the level of the project manager required is as large as possible. JATC has already figured out the answer and challenges you to do the same.

Input

The first line contains two integers n and q (2 ≤ n ≤ 100 000, 1 ≤ q ≤ 100 000) — the number of employees and the number of plans, respectively.

The second line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i ≤ n) meaning p_i is the direct boss of employee i.

It is guaranteed, that values p_i form a directed tree with the root of 1.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i<r_i ≤ n) — the range of the employees, involved in the corresponding plan.

Output

Print q lines, each containing two integers — the number of the employee which should be kicked from the corresponding plan and the maximum possible level of the project manager in that case.

If there are more than one way to choose that employee, print any of them.

Example

Input

11 5
1 1 3 3 3 4 2 7 7 6
4 6
4 8
1 11
9 11
8 11


Output

4 1
8 1
1 0
11 3
8 1

Note

In the example: 

<image> In the first query, we can choose whether 4 or 5 or 6 and the project manager will be 3.

In the second query, if we choose any employee other than the employee 8, the project manager will be 1. If we choose 8, the project manager will be 3. Since lv(3)=1 > lv(1)=0, choosing 8 is the best strategy.

In the third query, no matter how we choose the employee, the project manager will always be 1.

In the fourth query, if we choose 9 or 10 then the project manager will be 3. If we choose 11 then the project manager will be 7. Since lv(7)=3>lv(3)=1, we choose 11 as the answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int arr[200005];
long long int tree[800025];
void build(long long int node, long long int start, long long int endi) {
  if (start == endi) {
    tree[node] = arr[start];
  } else {
    long long int mid = (start + endi) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, endi);
    tree[node] = max(tree[2 * node], tree[2 * node + 1]);
  }
}
void update(long long int node, long long int start, long long int endi,
            long long int idx, long long int val) {
  if (start == endi) {
    arr[idx] = val;
    tree[node] = val;
  } else {
    long long int mid = (start + endi) / 2;
    if (start <= idx and idx <= mid) {
      update(2 * node, start, mid, idx, val);
    } else {
      update(2 * node + 1, mid + 1, endi, idx, val);
    }
    tree[node] = max(tree[2 * node], tree[2 * node + 1]);
  }
}
long long int query(long long int node, long long int start, long long int endi,
                    long long int l, long long int r) {
  if (r < start or endi < l) {
    return 0;
  }
  if (l <= start and endi <= r) {
    return tree[node];
  }
  long long int mid = (start + endi) / 2;
  long long int p1 = query(2 * node, start, mid, l, r);
  long long int p2 = query(2 * node + 1, mid + 1, endi, l, r);
  return max(p1, p2);
}
long long int arr2[200005];
long long int tree2[800025];
void build2(long long int node, long long int start, long long int endi) {
  if (start == endi) {
    tree2[node] = arr2[start];
  } else {
    long long int mid = (start + endi) / 2;
    build2(2 * node, start, mid);
    build2(2 * node + 1, mid + 1, endi);
    tree2[node] = min(tree2[2 * node], tree2[2 * node + 1]);
  }
}
void update2(long long int node, long long int start, long long int endi,
             long long int idx, long long int val) {
  if (start == endi) {
    arr2[idx] = val;
    tree2[node] = val;
  } else {
    long long int mid = (start + endi) / 2;
    if (start <= idx and idx <= mid) {
      update2(2 * node, start, mid, idx, val);
    } else {
      update2(2 * node + 1, mid + 1, endi, idx, val);
    }
    tree2[node] = min(tree2[2 * node], tree2[2 * node + 1]);
  }
}
long long int query2(long long int node, long long int start,
                     long long int endi, long long int l, long long int r) {
  if (r < start or endi < l) {
    return 10000000000;
  }
  if (l <= start and endi <= r) {
    return tree2[node];
  }
  long long int mid = (start + endi) / 2;
  long long int p1 = query2(2 * node, start, mid, l, r);
  long long int p2 = query2(2 * node + 1, mid + 1, endi, l, r);
  return min(p1, p2);
}
vector<long long int> graph[100005];
long long int t;
long long int node[1000005];
long long int dep[100005];
void dfs(long long int n) {
  arr2[n] = t;
  node[t] = n;
  t++;
  for (long long int i = 0; i < graph[n].size(); i++) {
    dfs(graph[n][i]);
  }
  arr[n] = t;
  node[t] = n;
  t++;
}
long long int lift[100005][22];
void dfs0(long long int n, long long int p) {
  lift[n][0] = p;
  for (long long int i = 1; i <= 17; i++) {
    lift[n][i] = lift[lift[n][i - 1]][i - 1];
  }
  for (long long int i = 0; i < graph[n].size(); i++) {
    dep[graph[n][i]] = dep[n] + 1;
    dfs0(graph[n][i], n);
  }
}
long long int tl, tr;
long long int dis;
long long int bin(long long int l, long long int r, long long int n) {
  if (l == r) {
    return l;
  }
  long long int x = (l + r) / 2;
  long long int y = n;
  for (long long int i = 17; i >= 0; i--) {
    if (x & (1 << i)) {
      y = lift[y][i];
    }
  }
  if ((arr2[y] <= tl) && (tr <= arr[y])) {
    return bin(l, x, n);
  } else {
    return bin(x + 1, r, n);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, q;
  t = 0;
  cin >> n >> q;
  for (long long int i = 2; i <= n; i++) {
    long long int z;
    cin >> z;
    graph[z].push_back(i);
  }
  dfs0(1, 0);
  dfs(1);
  build(1, 1, n);
  build2(1, 1, n);
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int mini = query2(1, 1, n, l, r);
    long long int maxi = query(1, 1, n, l, r);
    mini = node[mini];
    maxi = node[maxi];
    long long int ans = 0;
    long long int guy = 0;
    long long int kk;
    if (1) {
      long long int z1 = arr[maxi];
      long long int z2 = arr2[maxi];
      update(1, 1, n, maxi, 0);
      update2(1, 1, n, maxi, 1000000000);
      tl = query2(1, 1, n, l, r);
      tr = query(1, 1, n, l, r);
      update(1, 1, n, maxi, z1);
      update2(1, 1, n, maxi, z2);
      kk = l;
      if (l == maxi) {
        kk = r;
      }
      long long int v1 = bin(0, dep[kk], kk);
      v1 = dep[kk] - v1;
      if (ans <= v1) {
        ans = v1;
        guy = maxi;
      }
    }
    if (1) {
      maxi = mini;
      long long int z1 = arr[maxi];
      long long int z2 = arr2[maxi];
      update(1, 1, n, maxi, 0);
      update2(1, 1, n, maxi, 1000000000);
      tl = query2(1, 1, n, l, r);
      tr = query(1, 1, n, l, r);
      update(1, 1, n, maxi, z1);
      update2(1, 1, n, maxi, z2);
      kk = l;
      if (l == maxi) {
        kk = r;
      }
      long long int v1 = bin(0, dep[kk], kk);
      v1 = dep[kk] - v1;
      if (ans <= v1) {
        ans = v1;
        guy = maxi;
      }
    }
    cout << guy << " " << ans << "\n";
  }
}