Young Indian Grandmaster R Vaishali pulled off an upset win over former world champion Antaoneta Stefanova in the first leg of FIDE Women Speed Chess Championships but the country’s top player Koneru Humpy bowed out in the opener.

The Chennai-based Vaishali will take on International Master Munkhzul Turmunkh of Mongolia in the quarter-finals.

Vaishali registered a 6-5 win over Bulgarian opponent late on Wednesday after having beaten strong players like Valentina Gunina and Alina Kashlinskaya in the qualifying stages.

Reigning world rapid champion Humpy meanwhile went down 4.5-5.5 to her Vietnamese rival in the first round.

The scores were level at 3.5 points after the first two segments but the Vietnam player won the first two games of the final phase to secure a spot in the last eight.

Vaishali was delighted to put it across Stefanova.

“Yeah, it was great to play a former world champion and defeat her,” she told PTI.

“I was leading the match after one hour with 5.5-2.5 then in the bullet segment (but) internet got disconnected and I lost on time in an equal position,” she said after managing to pip her rival in a close finish.

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The Chennai player, sister of chess prodigy R Praggnanandhaa and Asian Blitz Championship winner in 2017, will meet Turmunkh in the last eight clash later on Thursday.

The Grand Prix will consist of four legs, with a total of 21 participants. Each of the 21 players participates in three out of four Grand Prix legs. Each GP is a 16-player knockout event.

In each Grand Prix leg, every player scores cumulative grand prix points according to the position in the final standings.

The two players who score the highest number of cumulative grand prix points in all three Grand Prix legs qualify for the Super Final to be held on July 20.