INF = 10000
N = gets.to_i
S = 1 << N
B = N.times.map { gets.split.map(&:to_i) }
dp = Array.new(N){ Array.new(S, INF) }

# dp[i][s]はsを使用して最後の本がiの時の栞の最大間隔

N.times do |i|
  dp[i][1 << i] = 0
end
(1 ... S).each do |s|
  used, unused = N.times.partition{ |i| (s & (1 << i)) > 0 }
  unused.each do |j|
    t = s | (1 << j)
    dp[j][t] = used.map{|i| [dp[i][s], B[i][1] - B[i][0] + B[j][0]].max }.min
  end
end
puts N.times.map{|i| dp[i][S - 1] }.min