Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 12;
const int Q = 1 << 16;
const int SQ = 1 << 8;
mt19937_64 rng(2020);
int n, q;
int a[N];
long long val[N];
int idx[N];
int sets[N];
int ans[Q];
int cnt;
pair<int, int> ops[2200005];
int merge(int u, int v) {
  if (u == -1) return v;
  if (v == -1) return u;
  ops[cnt++] = make_pair(u, v);
  return cnt - 1;
}
struct SegTree {
  long long h[N << 2];
  unordered_map<long long, int> elements[N << 2];
  void update(int node, int l, int r, int x) {
    if (l > x || r < x) return;
    h[node] ^= val[x];
    if (l == x && r == x) return;
    int mid = (l + r) >> 1;
    update(node * 2 + 1, l, mid + 0, x);
    update(node * 2 + 2, mid + 1, r, x);
  }
  int query(int node, int l, int r) {
    if (h[node] == 0) return -1;
    if (l == r) return idx[l];
    if (elements[node].find(h[node]) != elements[node].end()) {
      return elements[node][h[node]];
    }
    int mid = (l + r) >> 1;
    return elements[node][h[node]] = merge(query(node * 2 + 1, l, mid + 0),
                                           query(node * 2 + 2, mid + 1, r));
  }
  void update(int x) { update(0, 0, n - 1, x); }
  int query() { return query(0, 0, n - 1); }
} tree;
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    idx[--a[i]] = i;
    val[i] = rng();
  }
  vector<tuple<int, int, int>> queries(q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l, --r;
    queries[i] = make_tuple(l, r, i);
  }
  sort(queries.begin(), queries.end(),
       [](tuple<int, int, int> p, tuple<int, int, int> q) {
         return make_tuple(get<0>(p) / SQ, get<1>(p), get<2>(p)) <
                make_tuple(get<0>(q) / SQ, get<1>(q), get<2>(q));
       });
  cnt = n;
  int L = 0, R = -1;
  for (int i = 0; i < q; ++i) {
    int l, r, idx;
    tie(l, r, idx) = queries[i];
    while (L < l) tree.update(a[L++]);
    while (L > l) tree.update(a[--L]);
    while (R < r) tree.update(a[++R]);
    while (R > r) tree.update(a[R--]);
    ans[idx] = tree.query();
  }
  printf("%d\n", cnt);
  for (int i = n; i < cnt; ++i) {
    printf("%d %d\n", ops[i].first + 1, ops[i].second + 1);
  }
  for (int i = 0; i < q; ++i) printf("%d ", ans[i] + 1);
  printf("\n");
  return 0;
}