Statement : E. A Trivial String Problem

Define $f(t)$ as the maximum number of parts $t$ can be split into such that each part is a non-empty prefix of $t$. In other words, $f(t)$ is the maximum positive integer $k$ that satisfies the following condition:

• There exist $k$ strings $p_1,p_2,\ldots,p_k$, which are prefixes of $t$, such that $t=p_1+p_2+\ldots+p_k$. Here $+$ denotes the string concatenation.

You are given a string $s$ of length $n$, consisting of only lowercase English letters. Let $s[x..y]$ represent the substring$^{\text{∗}}$ of $s$ from the $x$-th to the $y$-th character (inclusive). You need to answer $q$ queries. The $i$-th query provides two integers $l_i$ and $r_i$, and you need to find

$$$ \sum_{j=l_i}^{r_i} f(s[l_i..j]). $

$^{\text{∗}}$A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$$$ by the deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $q$ ($1 \le n \le 10^6$, $1 \le q \le 100$).

The second line contains the string $s$ of length $n$, consisting of only lowercase English letters.

The $i$-th of the next $q$ lines contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$), representing the $i$-th query.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

For each query, output an integer representing the value of the expression.

Input

6
1 1
a
1 1
5 2
aaaaa
1 5
2 4
6 2
abcdef
1 6
3 5
6 3
abaaba
1 6
1 3
2 6
7 3
abcabca
1 7
2 7
4 7
8 3
aababaac
1 8
2 8
3 7

Output

1
15
6
6
3
14
4
7
12
9
5
13
14
7

Note

In the first test case, $f(\texttt{a})=1$.

In the second test case, $f(\texttt{a})+f(\texttt{aa})+f(\texttt{aaa})+f(\texttt{aaaa})+f(\texttt{aaaaa})=1+2+3+4+5=15$ and $f(\texttt{a})+f(\texttt{aa})+f(\texttt{aaa})=1+2+3=6$.

In the third test case, $f(\texttt{a})+f(\texttt{ab})+f(\texttt{abc})+f(\texttt{abcd})+f(\texttt{abcde})+f(\texttt{abcdef})=1+1+1+1+1+1=6$ and $f(\texttt{c})+f(\texttt{cd})+f(\texttt{cde})=1+1+1=3$.