#LCA(最近共通祖先)
#距離や間に挟まれているかも判定可能
import collections
import sys
sys.setrecursionlimit(10**7)

# N: 頂点数
# G[v]: 頂点vの子頂点 (親頂点は含む)

# Euler Tour の構築
N,Q = map(int, input().split())
N+=2
pair = [0]*(N)
S = list('('+input()+')')
G = [[] for _ in range(N)]
d = collections.deque()
dn = collections.deque()
d.append('x')
d.append('x')
for i in range(len(S)):
    if i>0 and S[i-1]=='(' and S[i]=='(':
        G[i].append(i-1)
        G[i-1].append(i)
    elif i>0 and S[i-1]==')' and S[i]=='(':
        G[dn[-1]].append(i)
        G[i].append(dn[-1])
    d.append(S[i])
    dn.append(i)
    if d[-2]=='(' and d[-1]==')':
        d.pop()
        d.pop()
        l = dn.pop()
        r = dn.pop()
        pair[l]=r
        pair[r]=l
        G[l].append(r)
        G[r].append(l)

S = []
F = [0]*N
depth = [0]*N

def dfs(v, d , p):
    F[v] = len(S)
    depth[v] = d
    S.append(v)
    for n in G[v]:
        if p!=n:
            dfs(n, d+1, v)
            S.append(v)
dfs(0, 0, -1)

# 存在しない範囲は深さが他よりも大きくなるようにする
INF = (10**6, 0)

# LCAを計算するクエリの前計算
M = 2*N
M0 = 2**(M-1).bit_length()
data = [INF]*(2*M0)
for i, v in enumerate(S):
    data[M0-1+i] = (depth[v], i)
for i in range(M0-2, -1, -1):
    data[i] = min(data[2*i+1], data[2*i+2])

# LCAの計算 (generatorで最小値を求める)
def _query(a, b):
    yield INF
    a += M0; b += M0
    while a < b:
        if b & 1:
            b -= 1
            yield data[b-1]
        if a & 1:
            yield data[a-1]
            a += 1
        a >>= 1; b >>= 1

# LCAの計算 (外から呼び出す関数)
def query(u, v):
    fu = F[u]; fv = F[v]
    if fu > fv:
        fu, fv = fv, fu
    return S[min(_query(fu, fv+1))[1]]

XY = [list(map(int, input().split())) for _ in range(Q)]

for x,y in XY:
    if x==y:
        print(min(x,pair[x]),max(x,pair[x]))
    else:
        l = query(x,y)
        if l == 0:
            print(-1)
        else:
            if l<pair[l]:
                print(l,pair[l])
            else:
                print(pair[l],l)