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 maxn = 5010;
const int maxm = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n, N, q, block, a[maxn], cnt, num, id[maxn], ans[maxm], belong[maxn];
struct node {
  int l, r, id;
  node(int a = 0, int b = 0, int c = 0) {
    l = a;
    r = b;
    id = c;
  }
  bool operator<(const node &a) const {
    return l < a.l || (l == a.l && r < a.r);
  }
};
vector<int> t[maxn], B[maxn];
vector<pair<int, int> > A[1000 * maxn];
pair<int, int> Q[maxm];
inline int Get(int x, int pos) {
  return lower_bound(B[x].begin(), B[x].end(), pos) - B[x].begin();
}
inline void Build(int x, set<node> Set) {
  B[x].clear();
  for (auto it : Set)
    if (B[x].empty() || B[x].back() != it.l) {
      B[x].push_back(it.l);
      A[x * N + Get(x, it.l)].clear();
    }
  for (auto it : Set)
    A[x * N + Get(x, it.l)].push_back(pair<int, int>(it.r, it.id));
}
inline int Find(int x, int l, int r) {
  auto L = lower_bound(B[x].begin(), B[x].end(), l);
  if (L == B[x].end()) return 0;
  auto R = upper_bound(A[x * N + Get(x, *L)].begin(),
                       A[x * N + Get(x, *L)].end(), pair<int, int>(r, INF));
  if (R == A[x * N + Get(x, *L)].begin()) return 0;
  --R;
  if (l <= *L && (*R).first <= r) return (*R).second;
  return 0;
}
inline set<node> Solve(int x, int l, int r) {
  set<node> ans;
  ans.clear();
  if (l > r) return ans;
  if (l == r) {
    ans.insert(node(id[l], id[l], id[l]));
    return ans;
  }
  int Mid = l + r >> 1;
  set<node> L = Solve(x, l, Mid), R = Solve(x, Mid + 1, r);
  for (auto it : L)
    if (it.l == it.r) ans.insert(it);
  for (auto it : R)
    if (it.l == it.r) ans.insert(it);
  Build(1, L);
  Build(2, R);
  for (int i = 0; i < t[x].size(); ++i)
    for (int j = i + 1; j < t[x].size(); ++j) {
      if (a[t[x][i]] < l || a[t[x][j]] < l || r < a[t[x][i]] || r < a[t[x][j]])
        continue;
      int posl = Find(1, t[x][i], t[x][j]), posr = Find(2, t[x][i], t[x][j]);
      if (!posl || !posr)
        ans.insert(node(t[x][i], t[x][j], posl + posr));
      else {
        printf("%d %d\n", posl, posr);
        ++num;
        --cnt;
        ans.insert(node(t[x][i], t[x][j], num));
      }
    }
  return ans;
}
set<node> Set[maxn];
int main() {
  scanf("%d%d", &n, &q);
  block = sqrt(q);
  if (n == 1) {
    puts("1");
    for (int i = (1), _end_ = (q); i <= _end_; ++i) printf("1 ");
    puts("");
    return 0;
  }
  N = block;
  for (int i = (1), _end_ = (n); i <= _end_; ++i)
    belong[i] = (i - 1) / block + 1;
  for (int i = (1), _end_ = (n); i <= _end_; ++i) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
    t[belong[a[i]]].push_back(i);
  }
  for (int i = (1), _end_ = (q); i <= _end_; ++i)
    scanf("%d%d", &Q[i].first, &Q[i].second);
  cnt = 2 * n * block + 2848;
  printf("%d\n", cnt);
  cnt -= n;
  num = n;
  for (int l = 1, r = min(n, block); l <= n; l += block, r = min(n, r + block))
    Set[belong[l]] = Solve(belong[l], l, r);
  for (int i = (1), _end_ = (belong[n]); i <= _end_; ++i) Build(i, Set[i]);
  for (int i = (1), _end_ = (q); i <= _end_; ++i) {
    int last = 0, l = Q[i].first, r = Q[i].second;
    for (int j = (1), _end_ = (belong[n]); j <= _end_; ++j) {
      int pos = Find(j, l, r);
      if (!pos) continue;
      if (last) {
        printf("%d %d\n", last, pos);
        ++num;
        --cnt;
        last = num;
      } else
        last = pos;
    }
    ans[i] = last;
  }
  while (cnt) {
    if (a[1] < a[2])
      puts("1 2");
    else
      puts("2 1");
    --cnt;
  }
  for (int i = (1), _end_ = (q); i <= _end_; ++i) printf("%d ", ans[i]);
  puts("");
  return 0;
}