根据题意，我们可以使用 复杂度求出以 为右边界的左边界在哪，然后把所有的询问离线，按照右端点从小到大排序，维护一个棵线段树，结点代表以 为右边界的最小区间长度，对于每个询问，只要把所有的右端点前的值更新到线段树中，再查询一次最小值即可知道答案。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e5 + 7;
const int maxm = 2000000 + 7;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int a[maxn];

int sum[maxn<<2];

void update(int M, int val, int l, int r, int rt){
  if(l == r){
    sum[rt] = val;
    return ;
  }
  int m = l + r >> 1;
  if(M <= m)  update(M, val, lson);
  else  update(M, val, rson);
  sum[rt] = min(sum[rt<<1], sum[rt<<1|1]);
}

int query(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return sum[rt];
  int ans = INF;
  int m = l + r >> 1;
  if(L <= m)  ans = query(L, R, lson);
  if(R > m)  ans = min(ans, query(L, R, rson));
  return ans;
}

struct Query{
  int id, l, r;
};
int ans[maxn];
Query q[maxn];

int main(){
  scanf("%d %d", &n, &m);
  ms(sum, INF);
  int sum = 0;
  map<int, int> mp;  mp[0] = 1;
  ms(a, INF);
  for(int i = 1; i <= n; ++i){
    int x;  scanf("%d", &x);
    sum ^= x;
    int y = mp[sum];
    if(y != 0)  a[i] = y;
    mp[sum] = i+1;
  }
  for(int i = 0; i < m; ++i){
    q[i].id = i;
    scanf("%d %d", &q[i].l, &q[i].r);
  }
  sort(q, q + m, [&](Query q1, Query q2){ return q1.r < q2.r; });
  int x = 1;
  for(int i = 0; i < m; ++i){
    while(x <= q[i].r){
      if(a[x] != INF)  update(a[x], x-a[x]+1, all);
      ++x;
    }
    int res = query(q[i].l, q[i].r, all);
    ans[q[i].id] = res == INF ? -1 : res;
  }
  for(int i = 0; i < m; ++i)  printf("%d\n", ans[i]);
  return 0;
}