Yet another ProblemHint 1
You are given an array aa of nn integers a1,a2,a3,…,an.
You have to answer qq independent queries, each consisting of two integers ll and rr.
- Consider the subarray a[l:r]a[l:r] == [al,al+1,…,ar][al,al+1,…,ar]. You can apply the following operation to the subarray any number of times (possibly zero)-
- Choose two integers LL, RR such that l≤L≤R≤r and R−L+1R−L+1 is odd.
- Replace each element in the subarray from LL to RR with the XOR of the elements in the subarray [L,R][L,R].
- The answer to the query is the minimum number of operations required to make all elements of the subarray a[l:r]a[l:r] equal to 00 or −1−1 if it is impossible to make all of them equal to 00.
You can find more details about XOR operation here.
Input
The first line contains two integers nn and qq (1≤n,q≤2⋅105)(1≤n,q≤2⋅105) — the length of the array aa and the number of queries.
The next line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai<230)(0≤ai<230) — the elements of the array aa.
The ii-th of the next qq lines contains two integers lili and riri (1≤li≤ri≤n)(1≤li≤ri≤n) — the description of the ii-th query.
Output
For each query, output a single integer — the answer to that query.
Example
input
7 6 3 0 3 3 1 2 3 3 4 4 6 3 7 5 6 1 6 2 2
output
-1 1 1 -1 2 0
Note
In the first query, l=3,r=4l=3,r=4, subarray = [3,3][3,3]. We can apply operation only to the subarrays of length 11, which won't change the array; hence it is impossible to make all elements equal to 00.
In the second query, l=4,r=6l=4,r=6, subarray = [3,1,2][3,1,2]. We can choose the whole subarray (L=4,R=6) and replace all elements by their XOR (3⊕1⊕2)=0(3⊕1⊕2)=0, making the subarray [0,0,0].
In the fifth query, l=1,r=6l=1,r=6, subarray = [3,0,3,3,1,2][3,0,3,3,1,2]. We can make the operations as follows:
- Choose L=4,R=6, making the subarray[3,0,3,0,0,0].
- Choose L=1,R=5, making the subarray[0,0,0,0,0,0].
对于一个区间，无论选哪对l, r做多少次异或操作，因为不能自己异或自己，所以异或和总是不变的。所以要想使一个区间都变为0，这个区间必须在初始时异或和就为0
异或和不变分析：
选择L, R， 设异或之后将L - R都替换为num，这是个奇数区间，奇数个num异或后还是num。
若是查询某个奇数区间：此区间全为0则 0 次操作
此区间异或和为0则 1 次操作
否则无法变为0
若是查询某个偶数区间：首先此区间要异或和全为0才可能产生答案
设此区间为l - r，因为只能选奇数区间异或，所以在l - r内必须存在一个奇数的异或和为0的子区间，先操作一次使这个子区间都为0，再选一个0和剩下的偶数个数异或，至此整个区间都为0
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